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Diary
Day 1
I was reading some typical examples of A Level experiments from several text books. I was stopped when I came to a few pages talking about refractive index. As in GCSE, I have learnt about the refractive index on air, water, glass and even diamond. But also being a chemist, having lots of experience in handling chemicals, making up solutions, what really interested me was that how will the refractive index of water changes if I add some substances into it? Does the amount matter? At that moment, I found myself in a position of wanting to discover or pursue what’s going to happen.
Day 2
Coming up with this idea, I started my planning on the procedure, my equipment list and the risk assessment. In order to make it nice and simple to start with, I planned to find the refractive index of pure water using the normal microscope. The microscope in Photo 1 was the one that I started with and was taken in the school. First of all, I put some pure water into a watch glass. Then I put a one-penny coin onto the bottom of the water. After that, I was going to put it onto the stage of the microscope. However it didn’t fix due to the short distance between the objective lens and the stage. Besides the magnification available wasn’t enough. After that, I had to get started on the travelling microscope. Unfortunately the set up of the travelling microscope that I found in my school was a little bit strange. The scale attached to the microscope was very accurate correct to 1mm. But the scale was perpendicular to the microscope. They were fixed and there were any ways for me to turn it round. So I wasn’t able to use the travelling microscope as I wouldn’t be able to take any measurements having the scale the other way round. I have taken two photos of the travelling microscope and they are shown in Photo 2 and 3.
Day 3
After that, I started using the binocular microscope shown in Photo 4. From those textbooks, I got some sort of idea that to find the refractive index of a solution, I will have to follow a few steps. First, set the microscope to some where in the scale as starting point. As the microscope doesn’t have a scale of its own, I made one up using graph paper. The microscope has to return back to where it started every time before adjusting it. Next fill a watch glass with solution. Then, put a coin onto the surface of the solution and allow it to sink to the bottom. Afterwards, place the watch glass under the microscope. Adjust the microscope until a clear picture of the coin is shown. Record the distance, r1, travelled by the microscope from where it starts to where it stops. Turn the microscope back to the starting point. Spread a small amount of lycopodium powder onto the surface of the solution. Again, adjust the microscope to have a clear picture of the coin. Record the distance, r2, travelled by the microscope from where it starts to where it stops this time. Finally remove the solution, just simply place the coin into the watch glass and find out the distance, r3, travelled by the microscope. I set the magnification to be 28. Now, when all the information needed is collected. There is a simple equation that can be used to find out the refractive index, n, of the solution. And it is:
n = real depth / apparent depth
= (r1-r3) / (r1-r2)
By knowing all these, I place 50 cm3 pure water onto a watch glass together with the one-penny coin with the magnification of 28, I found that r1-r3 = (2.1 ± 0.1) mm & r1-r2 = (1.1 ± 0.1) mm. The scale that I made was corrected to 0.1 mm, so when I recorded the values like r1-r3 & r1-r2 to 0.1 mm with a possible error of 0.1 mm. And these values give:
Upper boundary of n = 2.2 mm / 1 mm
= 2.2
Lower boundary of n = 2 mm / 1.2 mm
= 1.67
So the refractive index of water, n, is within the range of 1.67 – 2.2. In other words, the refractive index of water, n, is 1.94 ± 0.27.
Day 4
Even though the refractive index of water obtained is different from what I have learnt, but the method is giving me a reasonable answer. So I continued using the binocular microscope. As salt solutions and sugar solutions are easy enough to deal with, I decided to alter the concentration of these solutions and see what would happen. But before that, I have to know the maximum concentration of both salt and sugar solutions. I started with the salt first. First of all, I fill a 250 cm3 beaker with 200 cm3 of water. Then I kept pouring some salt in and at the same time I stirred it as well using a magnetic stirrer shown in Photo 5. I kept adding more and more salt until the solution contained more than enough salt to be saturated. And the weight of all the salt added has been recorded. Next I allowed the extra salt to settle. I weighed out the mass of a filter paper and I filtered off the salt solution using the filter paper and a filter funnel. I measured the mass of the filter paper together with the wasted salt. Errors have again been quoted. The numerical steps are shown below.
Volume of water = (200 ± 1) cm3
Mass of salt added = (90.00 ± 0.01) g
Mass of filter paper = (1.14 ± 0.01) g
Mass of filter paper and undissolved salt = (29.04 ± 0.01) g
Mass of undissolved salt = (27.90 ± 0.02) g
Mass of dissolved salt = (62.10 ± 0.03) g
Moles of salt = Mass of salt / Relative molecular mass of salt
Upper limit of the moles of salt = 62.13 g / 58.4 g mol-1
= 1.063870 mol
Lower limit of the moles of salt = 62.07 g / 58.4 g mol-1
= 1.062842 mol
Concentration of salt solution = Moles of salt / Volume of salt solution
Upper limit of maximum concentration of salt solution = 1.063870 mol / 199 cm3
= 0.005346 mol cm-3
= 0.5346 mol 100 cm-3
Lower limit of maximum concentration of salt solution = 1.062842 mol / 201 cm3
= 0.005288 mol cm-3
= 0.5288 mol 100 cm-3
So the solubility of salt in water measured in moles per 100 cm3 of water at 298K
= 0.5288 – 0.5346
= 0.5317 ± 0.0029
To make up a salt solution of 50 cm3,
Maximum moles of salt to be put in = Maximum solubility × Volume of water
Upper boundary of the maximum moles of salt to be put in
= 0.5346 mol 100 cm-3 × (50 ± 1) cm3
= (0.2673 ± 0.0053) mol
Lower boundary of the maximum moles of salt to be put in
= 0.5288 mol 100cm-3 × (50 ± 1) cm3
= (0.2644 ± 0.0053) mol
Maximum mass of salt to be put in = Maximum moles of salt × RMM of salt
Upper limit of maximum mass of salt to be put in
= (0.2673 ± 0.0053) mol × 58.4 g mol-1
= (15.6103 ± 0.3095) g
Lower limit of maximum mass of salt to be put in
= (0.2644 ± 0.0053) mol × 58.4 mol-1
= (15.4410 ± 0.3095) g
So by doing all the calculations above, I found the maximum mass of salt to be dissolved in 200 cm3 of water is (62.10 ± 0.03) g. And by using equations like:
Moles = Mass / Relative molecular mass
Concentration = Moles / Volume
I found the saturation of salt solution to be 0.5317 ± 0.0029 moles per 100 cm3.
Day 5
As I have decided to use 50 cm3 of water each time, after knowing that the saturation of salt solution, I worked out the amount of salt can be dissolved in 50 cm3 with the upper limit to be (15.6103 ± 0.3122) g and the lower limit to be (15.4410 ± 0.3088) g. So I tried to added different mass of salt into 50 cm3 within 15 g and allowed all the salt to dissolve with the help of a glass rod. Once all the solutions had been prepared, I actually carried what I wanted to find, i.e. the refractive of these solutions using the same method as I used in day 3 for water. The table of results of (r1-r3) & (r1-r2) values collected from those salt solutions are recorded in Appendix 1. I have also calculated the refractive index of different concentration of salt solutions with respect to the (r1-r3) & (r1-r2) found. All the calculations done on mole, concentration, refractive index etc following the same pattern of steps as shown in Day 3 &4.
Day 6
I used the same process for finding the saturation of salt solution as before to find the saturation for sugar solution. The numerical steps are shown below.
Volume of water = (200 ± 1) cm3
Mass of sugar added = (210.00 ± 0.01) g
Mass of filter paper = (0.21 ± 0.01) g
Mass of filter paper and undissolved sugar = (28.83 ± 0.01) g
Mass of undissolved sugar = (28.62 ± 0.02) g
Mass of dissolved sugar = (181.38 ± 0.03) g
Moles of sugar = Mass of sugar / Relative molecular mass of sugar
Upper limit of the moles of sugar = 181.41 g / 180.2 g mol-1
= 1.006715 mol
Lower limit of the mole of sugar = 181.35g / 180.2 g mol-1
= 1.006382 mol
Concentration of sugar solution = Moles of sugar / Volume of sugar solution
Upper limit of maximum concentration of sugar solution = 1.006715 mol / 199 cm3
= 0.005059 mol cm-3
= 0.5059 mol 100 cm-3
Lower limit of maximum concentration of sugar solution = 1.006382 mol / 200 cm3
= 0.004980 mol cm-3
= 0.4980 mol 100 cm-3
So the solubility of sugar in water measured in moles per 100 cm3 of water at 298K
= 0.4980 – 0.5059
= 0.5020
To make up a sugar solution of 50 cm3,
Maximum moles of sugar to be put in = Maximum solubility × Volume of water
Upper boundary of the maximum moles of sugar to be put in
= 0.5059 mol 100 cm-3 × (50 ± 1) cm3
= (0.2530 ± 0.0051) mol
Lower boundary of the maximum moles of sugar to be put in
= 0.5020 mol 100cm-3 × (50 ± 0.5) cm3
= (0.2490 ± 0.0050) mol
Maximum mass of sugar to be put in = Maximum moles of sugar × RMM of sugar
Upper limit of maximum mass of sugar to be put in
= (0.2530 ± 0.0051) mol × 180.2 g mol-1
= (45.5906 ± 0.9190) g
Lower limit of maximum mass of sugar to be put in
= (0.2490 ± 0.0050) mol × 180.2 mol-1
= (44.8698 ± 0.9010) g
And again, I also found out that the upper and lower limit of the mass of sugar to be put in 50 cm3 of water are (45.5906 ± 0.9190) g and (44.8698 ± 0.9010) g respectively. Following the same stages, I found the refractive index of sugar solution after that. Again, the (r1-r3) & (r1-r2) of those sugar solutions have been recorded in Appendix 2 together with the refractive index.
Day 7
However, I realised that there was a huge error when I was gathering the results. The magnification that I set was not sufficient enough. What this caused was that it gave a longer distance for the microscope to be moved and still got a clear image. This made the data collected inaccurate as I didn’t define where actually I should have stopped so my results were not reliable.
Day 8
Once I was aware of this situation, I tried to change the magnification to 56. Then I prepared all the salt and sugar solutions again. Unfortunately, it turned out that in all of the cases, the distances travelled to get a clear image of coin in just solution and solution with lycopodium powder are the same, which means that (r1-r2) was 0. At this point, my investigation was heading to nowhere. I quickly went back to those text books to see if there is anything that links to the concentration of solutions and I could investigate on.
Analysis
After collecting the refractive index of different concentration of salt solutions and sugar solutions, I have drawn out Fig.1 and Fig.2 showing how the refractive index changes when the concentrations of salt and sugar solutions are varied respectively.
In Fig.1, it gives a sort of pattern that the refractive index gradually increases as the concentration of salt solution increases apart from when the concentration is at (0.000343 ± 0.000010) mol cm-3. Basically, what the graph shows is that when the concentration of salt solution is increased, more salt is dissolved in the solution. This increases the density of the solution. This gradually increase in density means that the density of solution when comparing with the same medium, air, it is getting denser and denser. When the light passes through from air into the solution with the same angle of incidence, the light ray gets closer and closer to the normal. The angle of refraction is getting smaller and smaller. While sin i stays the same from concentration to concentration, when the angle of refraction gets smaller, sin r becomes smaller as well, which means that the ratio of sin i to sin r is effectively getting larger and larger. This contributes to the increase in refractive index as the concentration of salt solution is increased.
In Fig.2 shows nothing but points scattered all over the graph. Theoretically, when sugar solution gets more concentrated, the value of the density of the solution again should get higher; this in turns makes the light ray to bend towards normal having a smaller angle of refraction. Not surprisingly, the sin i / sin r ratio should get bigger and so as the refractive index. However, my experimental results don’t support this theory at all. There are a number of reasons for this and I am going to talk about them in the evaluation and conclusion. But what I want to mention now is that the theory describes about solutions in general but when sugar is dissolved in water, the solution is much thicker and more blur than the salt solutions. This may have contributed to the inaccuracy of my results.
Background
Conductance
The word conductance is the reciprocal of electrical resistance in a direct-current circuit, where resistance is the measure of a component’s opposition to the flow of electrical charge. Resistance is measured in ohms. At one time, the unit of conductance is measured in mho (the reverse of ohm, representing the reciprocal relationship between the two). The unit is renamed siemens after W. Siemens, who carried out extensive research into the behaviour of electricity.
There are two types of electrical conductors that their conductance can be measured and they are the electronic conductors and the electrolytic conductors. Electronic conductors include solid, molten metals, semiconductors and some salts while electrolytic conductors are electrolytes, molten salts and some ionic solids. I am not going to go into any deeper for electronic conductors as my investigation focuses on solutions, which is electrolytic conductors.
Conduction occurs in electrolytic conductors because there is a transport of ions from one part of the conductor to another, which means that both positive ions and negative ions migrate toward electrodes. There are 3 kinds of electrolytes in solutions and they are non-electrolytes (non-conductor), weak electrolytes (poor conductors) and strong electrolytes (good conductors). The conductance of a solution depends on a number of factors, it is
- directly proportional to the surface area of the electrodes
- inversely proportional to the distance between the electrode
(these two factors are held constant on the probe)
- directly proportional to the concentration of the ions in the solutions, the more ions it contains, the higher the conductance.
- directly on the mobility of the ions
- directly on the temperature (this is held constant during the experiment)
So what are left will going to be the main factors I am going to investigation on salt and sugar solutions, i.e. the concentration of the ions it contains and the mobility of its ions.
The most common method of measuring the conductance of a solution is by dipping a conductivity probe into the solutions. An electrical circuit is completed across the electrodes, which are on either side of the hole in the probe. A potential difference is applied to the two electrodes and current results, which is proportional to the conductance of the solution. Diagram 7 is an example of conduction with hydrochloric acid.
Diagram 7
Hydrochloric acid composes of hydrogen ions and chloride ions. Hydrogen ions are positively charged and chloride ions are negatively charge. When two electrodes connected to a power supply are placed into the solution, one of them has a positive charge and the other with negative charge. HCl molecules are separated into H+ and Cl- due to the attraction of opposite charge from the electrodes. The ions move towards the electrodes corresponding to the attraction and this creates a movement of ions in the solution. Because of this, the solution is able to conduct electricity.
Diary
Day 9
With my idea on refractive index failed, I then concentrated on the effect of changing the concentration of solutions on the conductivity. And I started with two electrodes connected to an ohmmeter. These two electrodes are then dipped into solutions of sugar or salt with different concentrations and the resistance of the solutions was recorded. By using the equation of , I worked out the conductance of each solution with the corresponding concentration. All the numerical values are shown in Appendix 3 with salt solutions and Appendix 4 with sugar solutions.
Day 10
Next, I prepared all the solutions again. What I did was to find out the conductance of these solutions using more specialised equipment, the conductivity meter shown in Photo 6. First, the conductivity meter was connected to a battery pack of 9V. Then I dipped the ‘dipping cell’ into the solution that I wanted to test for. By turning the knob to a specific conductance range, i.e. 10-5, 10-4, 10-3, 10-2, 10-1, 1 ohm-1, I then read off the conductance value of those solutions. This is shown in Appendix 5 with salt solutions and Appendix 6 with sugar solutions. Between each measurement, the ‘dipping cell’ has been washed.
Analysis
To analyse the rest of my investigation, after obtaining the values of conductance of salt and sugar solution with different concentrations, I have drawn out Fig. 3 – 6.
Fig.3 show that as the concentration of salt solution is increased, the conductance decreases. Both Fig.4 & Fig.6 show that the conductance of sugar solution decreases as its concentration is increased. These 3 graphs imply that as I increased the concentration of the solution, the number of ions or molecules per the same amount of volume is increased. This forces these ions or molecules to pack closer together, which means that they are not as free to move around. This means that the resistance of the solution increases. Because of the reciprocal relationship between resistance and conductance, as the resistance increases, the conductance decreases. So when I put the two electrodes connected to the ohmmeter or within the dipping cell into more concentrated solution, I got a smaller reading of conductance.
However, Fig.5 shows a completely different idea, which is that the conductance of salt solution increases when its concentration is increased. This can also be explained. As salt is made up of sodium ions and chloride ions. When salt is in solid state, sodium and chloride is held together by ionic bonds. But when it is dissolved in water, positive sodium ion, Na+, and negative chloride ion, Cl- are formed and they have the ability to move freely in the water. So more Na+ and Cl- ion are able to move towards the negative and positive electrode respectively.
Evaluation
In every experiment that we carry out, there ought to be some errors and limitations. And in my investigation, there are a number of these factors that might have contributed to the inaccuracy in the results I collected. And they are:
- errors or uncertainties in measurements and reading --- my whole investigation involved making up solutions, reading off values from scales of microscope ... where lots of these are only corrected to a certain level of accuracy. But I have calculated the uncertainties, i.e. the maximum and minimum errors in every bit of measurement. This is believed to be reasonable as I can include the error bars in my graphs and I can analyse from these graphs.
-
equipment not available --- this is a really big limitation to me as the text book suggests to use travelling microscope to find the refractive index. However, due to the travelling microscope having a scale the wrong way round, I have had to use binocular microscope. And it is because of this binocular microscope which gave me the values of R1 – R2 equalling to 0, the refractive index couldn’t be found.
- temperature --- temperature could be another factor that have affected the results obtained, as a higher temperature in some solutions would have caused more salt or sugar to be dissolved than in room temperature. I have tried to eliminate this error by carrying out all my experiments in the same laboratory, however, the temperature in the same laboratory can still differ from day to day even the difference shouldn’t be much.
Conclusion
To conclude, I wouldn’t say that I have actually proved anything or even I have deduced a law simply because what I found in both parts of my investigation, refractive index and conductance, don’t seem to match the theory and was totally unexpected. So far, I found out that for salt solution at a magnification of 28, as its concentration increases, the density of the solution is increasing; the angle of refraction of light ray bends more towards the normal, which gives a bigger refractive index. However, for sugar, there isn’t a pattern for its refractive index as its concentration is increased. Besides, as the magnification is adjusted to 56, the refractive index for both salt and sugar solution at any concentration is nothing. About the effect of varying the concentration on the conductance, sometimes increases the concentration results in having more freely charged ions to move towards electrodes giving a higher conductance and sometimes an increase in concentration results in reducing the ability for these ions to move around. And further more, even I have in fact discovered anything new or anything beyond the theory or anything that the theory couldn’t explain, without carrying out lots and lots of repeats, everything is still a mystery.
Bibliography
Reference
[1] Understanding Physics
For Advanced Level
Second Edition
Jim Breithaupt
[2] Physics
A Textbook For Advanced Level Students
2nd Edition
Tom Duncan
[3] Advanced Level
Practical Physics
Fourth Edition
M. Nelkon and J.M Ogborn
[4] Oxford Concise colour Science Dictionary
Oxford University Press 1997
Web page
[1]
[2]
[3] --- image search
Appendix
Appendix 1
See Fig.1
- mass is measured by a balance
- mole = mass / RMM
= (max mass / RMM + min mass / RMM) / 2 ± (max mass / RMM – min mass / RMM) / 2
= mean mole ± error in mole
= (max mole / min volume + min mole / max volume) / 2 ± (max mole / min volume - min mole / max volume) / 2
= mean conc. ± error in conc.
-
R1 – R3 & R1 – R2 are measured by the scale in binocular microscope
-
refractive index = (max R1 – R3 / min R1 – R2 + min R1 – R3 / max R1 – R2) / 2 ± (max R1 – R3 / min R1 – R2 – min R1 – R3 / max R1 – R2) / 2
= mean refractive index ± error in refractive index
Appendix 2
See Fig.2
- mass is measured using a balance
- mole = mass / RMM
= (max mass / RMM + min mass / RMM) / 2 ± (max mass / RMM – min mass / RMM) / 2
= mean mole ± error in mole
= (max mole / min volume + min mole / max volume) / 2 ± (max mole / min volume - min mole / max volume) / 2
= mean conc. ± error in conc.
-
R1 – R3 & R1 – R2 are measured the scale in binocular microscope
-
refractive index = (max R1 – R3 / min R1 – R2 + min R1 – R3 / max R1 – R2) / 2 ± (max R1 – R3 / min R1 – R2 – min R1 – R3 / max R1 – R2) / 2
= mean refractive index ± error in refractive index
Appendix 3
See Fig.3
- mass is measured by a balance
- mole = mass / RMM
= (max mass / RMM + min mass / RMM) / 2 ± (max mass / RMM – min mass / RMM) / 2
= mean mole ± error in mole
= (max mole / min volume + min mole / max volume) / 2 ± (max mole / min volume - min mole / max volume) / 2
= mean conc. ± error in conc.
- resistance is measured using a ohm metre
- conductance = (1 / min resistance + 1 / max resistance) / 2 ± (1 / min resistance – 1 / max resistance) / 2
= mean conductance ± error in conductance
Appendix 4
See Fig.4
- mass is measured by a balance
- mole = mass / RMM
= (max mass / RMM + min mass / RMM) / 2 ± (max mass / RMM – min mass / RMM) / 2
= mean mole ± error in mole
= (max mole / min volume + min mole / max volume) / 2 ± (max mole / min volume - min mole / max volume) / 2
= mean conc. ± error in conc.
- resistance is measured using a ohm metre
- conductance = (1 / min resistance + 1 / max resistance) / 2 ± (1 / min resistance – 1 / max resistance) / 2
= mean conductance ± error in conductance
Appendix 5
See Fig.5
- mass is measured by a balance
- mole = mass / RMM
= (max mass / RMM + min mass / RMM) / 2 ± (max mass / RMM – min mass / RMM) / 2
= mean mole ± error in mole
= (max mole / min volume + min mole / max volume) / 2 ± (max mole / min volume - min mole / max volume) / 2
= mean conc. ± error in conc.
- conductance is measured by a conductivity metre
Appendix 6
See Fig.6
- mass is measured by a balance
- mole = mass / RMM
= (max mass / RMM + min mass / RMM) / 2 ± (max mass / RMM – min mass / RMM) / 2
= mean mole ± error in mole
= (max mole / min volume + min mole / max volume) / 2 ± (max mole / min volume - min mole / max volume) / 2
= mean conc. ± error in conc.
- conductance is measured by a conductivity metre
